if we have a circuit with a bjt like this one enter image description here

and we know that the I is 1.1mA in 25 Celsius , with Vbe=680mV and Ie=1mA , and β=100 in 25 Celsius. If we know that the TC (thermal) of Vbe for constant Ie is -2mV/C how much would TC for Ve be? For 75 Celcius what is Ve?

I have calculated that for 25 Celsius Ib=9.9μΑ the current in R2 is 0.2mA going down and the current in R1 is 0.1mA going down. Also Vb=-13.6V and Ve=-14.28V . However I don't know how to calculate TC and the other question for 75 degrees. How could I do that?

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    \$\begingroup\$ This circuit is known as a Vbe multiplier. Is that enough of a hint for your homework question? \$\endgroup\$ Jun 6, 2019 at 22:30
  • \$\begingroup\$ @SpehroPefhany could you give me a good source for reading about this? because google does not help very much \$\endgroup\$
    – user170589
    Jun 6, 2019 at 22:42
  • \$\begingroup\$ Have you tried googling "Vbe multiplier"? \$\endgroup\$ Jun 6, 2019 at 22:51
  • \$\begingroup\$ @SpehroPefhany yes but nothing related to TC or any kind of formula to calculate that comes up \$\endgroup\$
    – user170589
    Jun 6, 2019 at 22:54
  • \$\begingroup\$ If you knew independently of this circuit what the TC of Vbe was, could you figure it out? \$\endgroup\$
    – TimWescott
    Jun 7, 2019 at 0:36

1 Answer 1


Your circuit has the basic structure of an NPN \$V_\text{BE}\$ multiplier. Typically, a \$V_\text{BE}\$ multiplier is operated with a current source (or sink, as in your case) and they provide a predictable voltage difference between the collector and emitter.

The basic idea is that the base-emitter junction will present a voltage difference of about \$V_\text{BE}\approx 700\:\text{mV}\$ at \$27^\circ\text{C}\$ (\$300\:\text{K}\$), which then appears across \$R_1\$. (In your case you are told a slightly different value for a slightly different temperature.) So \$V_{\text{R}_1}=V_\text{BE}\$. This implies a current in \$R_1\$ of \$I_{\text{R}_1}= \frac{V_\text{BE}}{R_1}\$, which must be supplied by a similar current in \$R_2\$.

The above paragraph I just wrote already conflicts with your writing:

the current in R2 is 0.2mA going down and the current in R1 is 0.1mA going down

I'm saying instead that the current in \$R_1\$ and \$R_2\$ is about the same (only slightly different because of the BJT's base recombination current.)

The voltage drop across \$R_2\$ is \$V_{\text{R}_2}= R_2\cdot \left(I_{\text{R}_1}+I_\text{B}\right)=V_\text{BE}\cdot\frac{R_2}{R_1}+R_2\cdot I_\text{B}\$ This difference across \$R_2\$ is added to \$V_{\text{R}_1}\$ so that \$V_\text{CE}= V_\text{BE}\cdot\left(1+\frac{R_2}{R_1}\right)+R_2\cdot I_\text{B}\$.

In your circuit's case, you were able to correctly estimate the base current as \$I_\text{B}\approx 9.9\:\mu\text{A}\$ based on being provided the emitter current, \$I_\text{E}=1\:\text{mA}\$ and the fact that \$\beta=100\$. But this base current must then be added to \$I_{\text{R}_1}= \frac{V_\text{BE}}{R_1}=100\:\mu\text{A}\$ so that \$I_{\text{R}_2}=I_{\text{R}_1}+I_\text{B}=109.9\:\mu\text{A}\$.

(I honestly don't know where you got your figure for \$R_2\$'s current.)

At this point, I think you can easily compute the voltage drop across \$R_2\$ and therefore the voltage drop across both resistors. This should provide a better answer for the simpler case prior to the temperature change.

(You may also will need to work out the value of \$I\$. And I think you do know how to do that. But whether or not it is important is a question that is addressed next.)

Once you have the above worked out, correctly, and you have the value of \$I\$ there is another problem to face, now. Does \$I\$ now stay exactly the same, when the temperature changes to \$75^\circ\text{C}\$? If so, then the emitter current will change and therefore so also will the base-emitter voltage and the base recombination current. But if the emitter current is instead said to remain constant then by implication the value of \$I\$ must instead change.

The latter choice would be more simply analyzed and wouldn't require changing the assumptions you were given. But I can't tell you which to hold as true since I don't know what's expected of you.

Assuming the simpler situation (where \$I\$ changes to accommodate a fixed value for \$I_\text{E}\$), then all you need to do is work out the change in \$V_\text{BE}\$ and from there work out the current in \$R_1\$, then the current in \$R_2\$ and then the voltage drop across \$R_2\$, etc.

The \$V_\text{BE}\$ multiplier has variations, as well.

The multiplier you have lacks any adjustment for the Early Effect, for example. A resistor is often added to the collector to provide an adjustment for that purpose to handle variations in the current passing through the multiplier.

Also, the tempco can be moved around, as well: either by providing the equivalent of a voltage source in the \$R_1\$ leg of the schematic or in the \$R_2\$ leg, depending on whether you want to increase or decrease the tempco.

A variable resistor is often added to the \$R_1\$ leg to make the multiplier adjustable, too.

Just be aware that you've got a simpler case and that there are other cases with additional parts to cope with what's important to the designer at the time.

  • \$\begingroup\$ I think it may be easier, instead of asking if \$I\$ will change with temperature, to ask if \$V_{BE}\$ will change with temperature. The temperature coefficient of the B-E voltage of a transistor is pretty well known, and pretty reliable (I used to work with equipment that was cryogenically cooled; we used \$V_{BE}\$ at a fixed current to measure temperature down to 77K). \$\endgroup\$
    – TimWescott
    Jun 7, 2019 at 17:37
  • \$\begingroup\$ @TimWescott Sorry I didn't reply earlier. (Out working on a pole barn in the rain.) I tried to stay clear of much mathematics and I think I really did (towards the end) stay focused on allowing the OP to work out to just focus on the VBE change, only. (I mentioned the I term, because I wanted the OP to be "knocked on the head" that the question itself could be viewed in different ways. (It's good to expand the mind just a little bit when the chance arises.) \$\endgroup\$
    – jonk
    Jun 7, 2019 at 19:43
  • \$\begingroup\$ @jonk Thank you for your answer, but do you know any formula to calculate the TC of vbe based on the changes of temperature? \$\endgroup\$
    – user170589
    Jun 7, 2019 at 22:11
  • \$\begingroup\$ @maverick98 You already wrote about how to do that. Do you recall the \$-2\:\frac{\text{mV}}{^\circ\text{C}}\$? Doesn't that already tell you enough? You are starting at \$25\:^\circ\text{C}\$ and the next question is for \$75\:^\circ\text{C}\$. What's the difference between these? How much will that impact your initial value of \$680\:\text{mV}\$ for the base-emitter voltage? \$\endgroup\$
    – jonk
    Jun 7, 2019 at 22:37
  • \$\begingroup\$ @jonk does β change with temperature? \$\endgroup\$
    – user170589
    Jun 12, 2019 at 15:28

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